WPE CYU 13

Derivation for Elastic Work and Sliding:

The process occurs in two phases as per the hint:

Phase 1: Initial Stretching (Force $0 \to F$)
Initially, a length $x_0$ is outside. The cord inside is held by friction. As we pull, the tension rises to $F$. Only the outside part ($x_0$) stretches effectively during this ramp-up.
The stiffness of the segment $x_0$ is $k_{out} = \frac{k l_0}{x_0}$.
Work done to stretch this part to force $F$: $$W_1 = \frac{1}{2} \frac{F^2}{k_{out}} = \frac{1}{2} \frac{F^2 x_0}{k l_0}$$ The extension achieved is $\delta_1 = \frac{F}{k_{out}} = \frac{F x_0}{k l_0}$.

Phase 2: Pulling Out (Constant Force $F$)
Once tension reaches $F$, the cord starts sliding out. We maintain force $F$ until the tail leaves the box.
Initial position of pull point: $x_0 + \delta_1$
Final position of pull point: When the cord is fully out, its total relaxed length is $l_0$. Under tension $F$, its total length is $l_0 + \Delta_{total}$, where $\Delta_{total} = F/k$.
Displacement of pull point during this phase: $$\Delta x = (l_0 + \frac{F}{k}) – (x_0 + \delta_1) = l_0 – x_0 + \frac{F}{k} – \frac{F x_0}{k l_0}$$ Work done in this phase ($W_2 = F \Delta x$): $$W_2 = F(l_0 – x_0) + \frac{F^2}{k} – \frac{F^2 x_0}{k l_0}$$

Total Work ($W = W_1 + W_2$): $$W = \frac{F^2 x_0}{2 k l_0} + F(l_0 – x_0) + \frac{F^2}{k} – \frac{F^2 x_0}{k l_0}$$ Combining the first and last terms ($\frac{1}{2}A – A = -\frac{1}{2}A$): $$W = F(l_0 – x_0) + \frac{F^2}{k} – \frac{F^2 x_0}{2 k l_0}$$ $$W = F(l_0 – x_0) + \frac{F^2}{k} \left( 1 – \frac{x_0}{2l_0} \right)$$